poj1094 拓扑 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 29744		Accepted: 10293

Description

An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.

Input

Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.

Output

For each problem instance, output consists of one line. This line should be one of the following three:

Sorted sequence determined after xxx relations: yyy...y.

Sorted sequence cannot be determined.

Inconsistency found after xxx relations.

where xxx is the number of relations processed at the time either a
sorted sequence is determined or an inconsistency is found, whichever
comes first, and yyy...y is the sorted, ascending sequence.

Sample Input

4 6
A<B
A<C
B<C
C<D
B<D
A<B
3 2
A<B
B<A
26 1
A<Z
0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.

Source

East Central North America 2001

 

 

代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int map[100][100];
int m,n;
int tindegree[100],indegree[100];
char str[5];
char s[39];
int toposort(){
     bool flag=true;
     memset(tindegree,0,sizeof(tindegree));
     memset(s,0,sizeof(s));
     for(int i=1;i<=n;i++){
     tindegree[i]=indegree[i];
     }
    for(int i=1;i<=n;i++){
        int sum=0,k;
           for(int j=1;j<=n;j++){
               if(!tindegree[j]){
                    k=j;
                    sum++;
               }
           }
           if(sum==0){///入度为0，则所剩图为一个环，无法判断了。直接可以返回，
              return -1;
           }
           if(sum>1){///当出现入度为0的点有多个时候，可能判断不了，但是也不能直接返回0，因为带环循环会优先于
                   ///无法判断这种情况，所以需要继续执行该函数，看还有没有死循环为环的情况
                flag=false;
           }
        s[i-1]=k+'A'-1;
        tindegree[k]--;
        for(int z=1;z<=n;z++){
            if(map[k][z]){
                 tindegree[z]--;
            }
        }
 
    }
    s[n]='\0';///s字符串结束标志
    if(flag==false)
    return 0;
    return 1;
}
int main(){
     while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0&&m==0)
           break;
          memset(map,0,sizeof(map));
          memset(indegree,0,sizeof(indegree));
          memset(str,0,sizeof(str));
          memset(s,0,sizeof(s));
          int ans=2;
          int temp;
           bool flag=true;
          for(int i=1;i<=m;i++){
              scanf("%s",str);
              if(flag==false)
              continue;
              int u=str[0]-'A'+1;
              int v=str[2]-'A'+1;
              if(!map[u][v]){
                 map[u][v]=1;
                 indegree[v]++;
              }
             ans=toposort();
             if(ans==-1||ans==1){///此判断语句特别注意，只能记录状态，不能退出，需要继续读边
                 temp=i;
                 flag=false;
             }
          }
          if(ans==1)
          printf("Sorted sequence determined after %d relations: %s.\n",temp,s);
          else  if(ans==-1)
          printf("Inconsistency found after %d relations.\n",temp);
          else  if(ans==2)
          printf("Sorted sequence cannot be determined.\n");
 
     }
    return 0;
}